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I have a few doubts in the proof of existence of free groups given by Clara Loeh in page-22,23 of her Geometric Group theory book.

Theorem 2.2.7: Let S be a set. Then there exists a group freely generated by S. (By the previous proposition, this group is unique up to isomorphism.)

  1. When the equivalence is defined in page-22, is it implicitly assumed that $ s$ and $ \hat{s}$ are inverses?

$$\forall_{x,y \in A^*} \forall_{s \in S} xs \hat{s}y \sim y xy$$ $$\forall_{x,y \in A^*} \forall_{s \in S} x \hat{s} s y \sim xy$$

  1. In page-23, Clara defines a function which help us define inverse for general words. Here is what she gave exactly:

Inductively (over the length of sequences), we define a map $I:A^* \to A^* $ by $I(\epsilon) := \epsilon$ and

  1. $I(sx) :=I(x) \hat{s}$

  2. $I( \hat{s} x) : = I(x) s$

In the above why did she write 1. and 2.? Shouldn't just one of the equations be enough? My thinking is that in $A^*$ , $s$ and $\hat{s}$ are just elements in the set $A^*$.

Secondly, I am not sure if I understand this function in itself. Suppose I have a word $abc$ made up of letters $a,b,c$ then would it be correct to write $I(abc)= I(bc) \hat{a}$..? and then push it down till $I(abc)=\hat{c} \hat{b} \hat{a}$?

  1. In page-23 (again), I am confused at the part where she defines $\varphi^*$ which map from the set to words to a group. The idea can be understood by the diagram: enter image description here

For every group G and every map$ \varphi: S −→ G$ there is a unique group homomorphism $\varphi: F(S) −→ G$ such that $\overline{\varphi} ◦ i = \varphi$. Given $\varphi$, we construct a map

$\phi^*: A^* \to G$

Inductively by:

  1. $ \epsilon \to e$

  2. $sx \to \varphi(s) \cdot \varphi^*(x)$

  3. $ \hat{s} x \to ( \varphi(s))^{-1} \cdot \varphi^*(x)$

Again I am confused, why we need both two and three..? I think one is enough for similar reason said before.


Note:

  1. Free groups are defined by their universal property in the book.
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For your first question: the proof of theorem $2.2.7$ explicitly explains what $\hat{s}$ is: it's an additional disjoint copy of $S$ whose elements play the roles of inverses. In fact, it's the very first paragraph of the proof.

For your second question: since $s\in S$ and $\hat s \in S$ are different (and belong to different, disjoint copies of $S$) we need to define how they both operate on the group element $x$, which is why both points $1$ and $2$ are given. You can indeed go from $I(abc)$ to $\hat{c}\hat b\hat a$ and this is also the expected right-inverse of the word $abc$ in a group.

And for your last question the answer is again that this explicitly connects the nominal inverse $\hat s$ with the inverse of the homomorphism $\varphi$ so both need to be defined.

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  • $\begingroup$ If the group is not commutative then knowing what $x\hat s$ is tells you nothing about $\hat s x$. So you have to define both terms separately. Nominal inverse means "that thing we are calling an inverse" because: $S$ is set and sets don't have inverses. But if we can treat $S$ as a group, then it does, because groups are required to have inverses. Since groups are also closed, all the inverses $\hat s$ of elements $s\in S$ must also be in $S$, so we can take a second copy of $S$ and treat them as in the inverses. We're calling them inverses, but at this stage we haven't shown they are $\endgroup$
    – postmortes
    Apr 12 at 11:38
  • $\begingroup$ I don't really understand what you're writing there I'm afraid. $\hat s x$ and $x \hat s$ are different things, as Ms. Löh defines, and are on equal footing inasmuch as they are both words. $\endgroup$
    – postmortes
    Apr 12 at 11:55
  • $\begingroup$ Ah ok! I'm sorry, that's completely my fault; points one and two are not about commutativity (I won't bore you with the connection that made me write that); I've updated the answer to reflect that. The issue is that $s$ and $\hat s$ belong to different copies of $S$ so we need to specify how they both behave with respect to $I$. $\endgroup$
    – postmortes
    Apr 12 at 14:23
  • $\begingroup$ Could you explain right side of third eqtn? I am confused at $( \varphi(s))^{-1}$ $\endgroup$ Apr 12 at 14:41
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    $\begingroup$ If you have more questions you should really ask them as a new question and not keep adding them in comments. However: the point of $3.$ is to establish that $\phi ^*$ is a homomorphism and that it interacts with $\hat s$ by inverses (because we intend $\hat s$ to be the inverse of $s$). From $\epsilon = \phi(\varepsilon) = \phi(\hat s s) = \phi(\hat s) \phi(s) \Rightarrow (\phi(s))^{-1} = \phi(\hat s)$ (writing the homomorphism multiplicatively) being what we want we obtain $3$. $\endgroup$
    – postmortes
    Apr 13 at 6:46

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